Recent content by lukka98

Ok, maybe my question was misunderstood, but ##X(s) = \frac{1}{1 + (1+s)^2}##, so for s = -1, I have X(s) = 1; but ##\int_{0}^{\infty} sin(t) dt## ,that is for s = -1, is indefinite. And also the graphics solution, looks like X(s) = 1 for s = -1.

I have a f(t) that is, e^(-t) *sin(t), now I calculate the Laplace transformation, that is: X(s) = 1 / ( 1 + ( 1 + s)^2 ) (excuse me but Latex seems not run ). Now I imagine the plane with Re(s), Im(s) and the magnitude of X(s). If i take Re(s) = -1 and Im(s) = 0, I believe I have X(s) = 1 ( s...

In the fermi gas model, there is assumption that there is a 3D potential well, but there is "energetic degeneration" for each three index "nx, ny, nz". Now the problem is with that image, if there is degeration, for some level En there may be 10 distinctive state with same energy, so there is 20...

Ok, what disturbed me is that the voltage across capacitor is equal at every time to the e.m.f. generated by inductor... or i believe so. I see like there are two generator equal in magnitude but opposite in direction, or the voltage across the inductor is like a "drop" of potential?

I solve it using Laplace transformation, and "analyzing" the T(s), so at the end I have: ##V_c (t) = \frac{Q}{C} *cos(\omega_0 t)## so the graph is a cos with angular frequency ##\omega_0## of (1/LC)^(1/2). V_c(t) = V_L(t), and ##\frac{dI}{dt} = \frac{Q}{LC}## with Q the initial charge of...

Without mathematical formulas, but only with a "Physical intuitive meaning", why if at t=0, I have a charged capacitor, and I connect it through a wire ,forming a closed path, to a inductor the current increasing with time and his derivative decreasing? To me seems like the inductor oppose "to...

there is a thing that I have difficult to accept: If, instead of a person I take an atom ( only for more real example) with the same 0.5 prob. to decay per day. After 1 million of year of him "creation" there is a probability of 10^-50 that is already live, I know every day it survive with 0.5...

My situation was ideal, just for this question.

Ok, but every day I have always a probability of 0.5 to survive, is correct? and for example each 4 days i have always a probability of 0.0625 to survive, now or after 1 year?

So what I calculate is the probability to be live "2 days consecutively?"

For example, If I have a constant probability of 50% to death per day (sorry for the macabre example but I find it good); Every day I wake up, I can say:"Today, at the end of the day, I have a probability of 0.5 to be live( or to be dead)!". Now, one can calculate the probability that after...

In last, then I stop, so ##\lambda ## is equal to the "probability per unit time that the particle decays" only for "small" ##\lambda##?

Excuse me, If I can I have a thing, an example: ## ^{215}Po ## has a ## \lambda = 3.9 * 10^{2} s^{-1}##; If I take the variation, according to equation that describe the decay, ## \Delta N(t) = - \lambda N(t) \Delta t ##, for example having ##N(t) = 10 ##. I have ##\Delta N(t) = -3900 ## for...

So for example if ##\tau## is 1 years, ##\frac{1}{\lambda}## is 1, and the probability to having a decay in 1 year is ... 1? If i would calculate the probability of a decay of a particle of average lifetime of 1 years, after 1 years, how I do this? I can't be 1 I think.

this answer give rise me a question ( maybe stupid): ##\lambda## is the probability per unit time that a particle decay, but if a particle have an average lifetime of ##\[ 10^{-20} s]\## so the decay probability for unit time ( i think per second..?) is greater than 1, but this is nosense, so...